An example that shows weak law of large number fails and a question about it.

$X_n$ is a sequence of independent random variables with $\mathbb{P}[X_n=n^2-1]=n^{-2},\mathbb{P}[X_n=-1]=1-n^{-2}$ and $Var[X_n]$ is unbounded. Set $S_n=X_1+...+X_n$. Prove that $\frac{S_n}{n} \rightarrow -1$ in probability.

Till now, I have proved that $\mathbb{E}[X_n]=0$. But then I am confused, no matter by weak law of large numbers or strong law of large numbers, this doesn't make sense to me that $\frac{S_n}{n}$ converges to -1. Can anyone explain to me what happened? Thanks!

• Doesn't the weak law only hold for i.i.d random variables? (ie, they need to be identically distributed). – Andrew D Apr 11 '13 at 23:05
• I don't think so. Weak law only need those random variables uncorrelated and the variance is bounded. – Cancan Apr 11 '13 at 23:06
• Going off MathWorld (mathworld.wolfram.com/WeakLawofLargeNumbers.html) seems to suggest otherwise; having had a quick look around elsewhere, it appears that the strong law holds for non-identically distributed random variables provided they follow some stricter conditions: (mathworld.wolfram.com/StrongLawofLargeNumbers.html) – Andrew D Apr 11 '13 at 23:10
• This is is also true, because strong law only requires them to be independent but not necessarily identically distributed. :) – Cancan Apr 11 '13 at 23:12

The series $\sum\limits_nP[X_n\ne-1]$ converges hence $X_n=-1$ for every $n$ large enough, almost surely, say for every $n\geqslant N$. Thus, $S_n=S_N+N-n$ for every $n\geqslant N$, in particular $S_n/n\to-1$ almost surely (hence also in probability). The independence hypothesis is not needed.
• And in addition, why do you write $S_n=S_N+N-n$. Why do you get the $N-n$ in the equation? :) – Cancan Apr 11 '13 at 23:26
• About $N-n$: say, if $X_k=-1$ for every $k\geqslant N$, what is $S_n-S_N$ for $n\geqslant N$? – Did Apr 11 '13 at 23:31
• Sorry, I thought the equation $S_n= S_N+N-n$ again and again and I couldn't figure out why is so? Till now can conclude that $X_n \displaystyle \rightarrow^{a.s.} -1$ but how did you get the conclusion from here? still not clear to me. – Cancan Apr 11 '13 at 23:38